Solution:

Let us solve this problem with variables to find the general nature. Let us assume that there are x black chocolates and y white chocolates. Three things can happen when you take out two chocolates.
1) You take out two black chocolates.
Then, you will be replacing one black chocolate and so the jar will have x+1 black chocolates and y-2 white chocolates.
2) You take out two white chocolates.
Then, you will be replacing 1 black chocolate and so the jar will have x-1 black chocolates and y white chocolates.
3) You take out 1 black chocolate and 1 white chocolate. Then, you will be replacing a white chocolate and so the jar will have x-1 black chocolates and y white chocolates.
Thus at every step, you are wither removing a black chocolate or replacing two white chocolates by one black chocolate irreversibly till there is just one chocolate left. Therefore each white chocolate is equivalent to two black chocolates.

This, if the number of white chocolates are odd in number, the last one remaining will be white chocolate and if the number of white chocolates are even in number, the last one remaining will be black chocolate.

Solution:

74658

* 8 + 6 =14
* 5 - 4 = 1
* 7 = 2*4 - 1
* 4 + 6 = 10
* 7+4+6+5+8 = 30

Solution:

On Your lap.

Solution:

They are matched as shown in the picture below.

Solution:

10

Explanation:
Equation 1 + 9 + 11 = 1 can be derived from
One (o) + nine (n) + eleven (e) = one => 1

Similarly for equation,
12 + 11 + 9
Twelve (t) + eleven (e) + nine (n) => ten (10)

Solution:

Now there can be two solutions to the given situation because there can be two situations:
Suppose if B and C have same colors, A will know that his color is the other one and he will be able to speak up the first.
Now if B and C do not have same colors, A will stay silent. This will tell B that his and C's colors are different. Thus he will speak up first.

So either A or B will speak up first depending on the situation.

Solution:

3,3,8

Lets break it down. The product of their ages is 72. So what are the possible choices?

2, 2, 18 sum(2, 2, 18) = 22
2, 4, 9 sum(2, 4, 9) = 15
2, 6, 6 sum(2, 6, 6) = 14
2, 3, 12 sum(2, 3, 12) = 17
3, 4, 6 sum(3, 4, 6) = 13
3, 3, 8 sum(3, 3, 8 ) = 14
1, 8, 9 sum(1,8,9) = 18
1, 3, 24 sum(1, 3, 24) = 28
1, 4, 18 sum(1, 4, 18) = 23
1, 2, 36 sum(1, 2, 36) = 39
1, 6, 12 sum(1, 6, 12) = 19

The sum of their ages is the same as your birth date. That could be anything from 1 to 31 but the fact that Torres was unable to find out the ages, it means there are two or more combinations with the same sum. From the choices above, only two of them are possible now.

2, 6, 6 sum(2, 6, 6) = 14
3, 3, 8 sum(3, 3, 8 ) = 14

Since the eldest kid is taking piano lessons, we can eliminate combination 1 since there are two eldest ones. The answer is 3, 3 and 8.

Solution:

36 meters.

100/v1 = 80/v2
100/v2 = 80/v3

20+(20/v2)*v3=20+16 = 36

Solution:

So that the following plan can be followed, let us number the coins from 1 to 12. For the first weighing let us put on the left pan coins 1,2,3,4 and on the right pan coins 5,6,7,8.

There are two possibilities. Either they balance, or they don't. If they balance, then the different coin is in the group 9,10,11,12. So for our second one possibility is to weigh 9,10,11 against 1,2,3

(1) They balance, in which case you know 12 is the different coin, and you just weigh it against any other to determine whether it is heavy or light.
(2) 9,10,11 is heavy. In this case, you know that the different coin is 9, 10, or 11, and that that coin is heavy. Simply weigh 9 against 10; if they balance, 11 is the heavy coin. If not, the heavier one is the heavy coin.
(3) 9,10,11 is light. Proceed as in the step above, but the coin you're looking for is the light one.

That was the easy part.

What if the first weighing 1,2,3,4 vs 5,6,7,8 does not balance? Then any one of these coins could be the different coin. Now, in order to proceed, we must keep track of which side is heavy for each of the following weighings.

Suppose that 5,6,7,8 is the heavy side. We now weigh 1,5,6 against 2,7,8. If they balance, then the different coin is either 3 or 4. Weigh 4 against 9, a known good coin. If they balance then the different coin is 3, otherwise it is 4. The direction of the tilts can tell us whwther the offending coin is heavier or lighter.

Now, if 1,5,6 vs 2,7,8 does not balance, and 2,7,8 is the heavy side, then either 7 or 8 is a different, heavy coin, or 1 is a different, light coin.

For the third weighing, weigh 7 against 8. Whichever side is heavy is the different coin. If they balance, then 1 is the different coin. Should the weighing of 1,5, 6 vs 2,7,8 show 1,5,6 to be the heavy side, then either 5 or 6 is a different heavy coin or 2 is a light different coin. Weigh 5 against 6. The heavier one is the different coin. If they balance, then 2 is a different light coin.

Solution:

Both will rotate at Same speed.

Explanation:
Gears of the same size will always rotate at constant speeds despite any smaller or larger gears in between them.

Solution:

Numbers of the first cube:
1, 2, 4, 0, 5 and 6.

Numbers on the second cube:
3, 1, 2, 7, 8 and 0.

Now you can pick any date and you will be able to show it using these cubes. Remember that you can always use 6 as 9 by inverting it.

This problem has another solution and we leave it for you to deduce.